Question

1/2x-1/4y+z=-51/2x-1/3y-1/3z=0x-y+2/3z=-20/3the solution set or no solution inconsistent/dependent or infinite answers inconsistent/dependent

Answer 1

The given system of equations is

$\begin{gathered} \frac{1}{2}x-\frac{1}{4}y+z=-5 \\ \frac{1}{2}x-\frac{1}{3}y-\frac{1}{3}z=0 \\ x-y+\frac{2}{3}z=-\frac{20}{3} \end{gathered}$

First, let's multiply the second equation by -1 to then add it with the first equation.

$\begin{gathered} \frac{1}{2}x-\frac{1}{4}y+z=-5 \\ -\frac{1}{2}x+\frac{1}{3}y+\frac{1}{3}z=0\rightarrow\frac{1}{2}x-\frac{1}{2}x-\frac{1}{4}y+\frac{1}{3}y+z+\frac{1}{3}z=-5+0 \\ \end{gathered}$

Now, let's reduce like terms

$\begin{gathered} 0x+\frac{-3y+4y}{12}+\frac{3z+z}{3}=-5 \\ \frac{y}{12}+\frac{4z}{3}=-5 \end{gathered}$

As you can observe, we've got an equation with two variables, y, and z. Now, we have to find another equation with the same variables. Let's multiply the third equation by -1/2 to then add it with the first equation.

$\begin{gathered} -\frac{1}{2}x+\frac{1}{2}y-\frac{2}{6}z=\frac{20}{6} \\ \frac{1}{2}x-\frac{1}{4}y+z=-5\rightarrow\frac{1}{2}x-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{4}y+z-\frac{1}{3}z=\frac{10}{3}-5 \end{gathered}$

Now, we reduce like terms

$\begin{gathered} 0x+\frac{1}{4}y+\frac{2}{3}z=\frac{10-15}{3} \\ \frac{1}{4}y+\frac{2}{3}z=\frac{-5}{3} \end{gathered}$

Then, we form a new system of two equations with the new equations we found

$\begin{gathered} \frac{y}{12}+\frac{4z}{3}=-5 \\ \frac{y}{4}+\frac{2z}{3}=-\frac{5}{3} \end{gathered}$

At this point, let's solve this new system. We have to multiply the second equation by -1/3 to then add them

$\begin{cases}\frac{y}{12}+\frac{4z}{3}=-5 \\ -\frac{y}{12}-\frac{2z}{9}=\frac{5}{9}\end{cases}\rightarrow\frac{4z}{3}-\frac{2z}{9}=-5+\frac{5}{9}$

Let's solve for z

$\begin{gathered} \frac{36z-6z}{27}=\frac{-45+5}{9} \\ \frac{30z}{27}=\frac{-40}{9} \\ z=\frac{-50\cdot27}{9\cdot30} \\ z=\frac{-1080}{270} \\ z=-4 \end{gathered}$

Then, we find the value of y using one of the equations in the last system.

$\begin{gathered} \frac{y}{4}+\frac{2z}{3}=-\frac{5}{3} \\ \frac{y}{4}+\frac{2\cdot(-4)}{3}=-\frac{5}{3} \\ \frac{y}{4}-\frac{8}{3}=-\frac{5}{3} \\ \frac{y}{4}=-\frac{5}{3}+\frac{8}{3} \\ \frac{y}{4}=\frac{3}{3} \\ y=1\cdot4 \\ y=4 \end{gathered}$

At last, we use one of the initial equations to find the value of x.

$\begin{gathered} \frac{1}{2}x-\frac{1}{4}y+z=-5 \\ \frac{1}{2}x-\frac{1}{4}\cdot4+(-4)=-5 \\ \frac{1}{2}x-1-4=-5 \\ \frac{1}{2}x-5=-5 \\ \frac{1}{2}x=-5+5 \\ x=2\cdot0 \\ x=0 \end{gathered}$

Therefore, the system has the solution set (0, 4, -4), that is, x = 0, y = 4, and z = -4.